Find the area between the cylinder $z^2+y^2=r^2$ and two planes I'm having trouble with this problem:

Find the surface area between the top of $z^2+y^2=r^2$ between $z=ax$ and $z=bx$ (consider $a \gt  b  \gt 0$).

I think I must find the area between the cylinder and the two planes where $z \ge0 $, and depending on the values of $r, a,b$ it looks like:

I do not know how to start; should I parametrize the intersection and then integrate?
Any ideas?
 A: There is a symmetry.  
There is as much area between the planes on one side of the line intersection of the planes ($x=z = 0$)as on the other side of the line.
The total area then is double the area in 1/2 the cylinder. 
Lets rewrite the planes as:
$x = z/a$ and $x = z/b$
covert to polar:
Surface area  = $\int_0^{\pi}\int_{(r/b) \sin\theta}^{(r/a) \sin\theta} r\, dx d\theta$
A: We are talking here about the half-cylinder in $x$-direction, intersecting the $(y,z)$-plane in the semicircle $$\gamma:\qquad y=\cos\phi,\quad z=\sin\phi\qquad(0\leq\phi\leq\pi)\ .$$
Through each point  $(y,z)\in \gamma$ there is an infinite stalk in $x$-direction, but only the part
$${z\over a}\leq x\leq{z\over b}$$ of length
$$\ell(\phi):={z\over b}-{z\over a}={a-b\over ab}\>r\sin\phi$$
of this stalk belongs to the surface $S$ in question. As $ds=r\>d\phi$ along $\gamma$ the total area of $S$ computes to
$${\rm area}(S)=\int_0^\pi \ell(\phi)\>r\>d\phi={a-b\over ab}\>r^2\int_0^\pi\sin\phi\>d\phi=2r^2{a-b\over ab}\ .$$
A: According to $z=ax$ and $z=bx$ and that $a>b>0$, you see that $z/a\leq x\leq z/b$. In fact, this shows the variation of $x$. You can easily find the variations of $y$ and $z$ using polar coordinates. Since $z\geq 0$, so we have $0\leq \theta \leq \pi$ and $0\leq R\leq r$ which $(R, \theta)$ are polar variables.
